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Calculate [tex]\(\Delta H\)[/tex] for the reaction:
[tex]\[
N_2H_4(l) + CH_4O(l) \rightarrow CH_2O(g) + N_2(g) + 3H_2(g)
\][/tex]
using the following reactions and their [tex]\(\Delta H\)[/tex] values:
[tex]\[
\begin{array}{ll}
2NH_3(g) \rightarrow N_2H_4(l) + H_2(g) & \Delta H = 22.5 \, \text{kJ} \\
\hline
2NH_3(g) \rightarrow N_2(g) + 3H_2(g) & \Delta H = 57.5 \, \text{kJ} \\
\hline
CH_2O(g) + H_2(g) \rightarrow CH_4O(l) & \Delta H = -81.2 \, \text{kJ}
\end{array}
\][/tex]



Answer :

GinnyAnswer
To calculate the enthalpy change (ΔH) for the reaction:

[tex]\[ \text{N}_2\text{H}_4(l) + \text{CH}_4\text{O}(l) \rightarrow \text{CH}_2\text{O}(g) + \text{N}_2(g) + 3\text{H}_2(g) \][/tex]

we are given the following reactions and their enthalpy changes:

1. [tex]\[ 2\text{NH}_3(g) \rightarrow \text{N}_2\text{H}_4(l) + \text{H}_2(g) \quad \Delta H = 22.5 \text{ kJ} \][/tex]
2. [tex]\[ 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta H = 57.5 \text{ kJ} \][/tex]
3. [tex]\[ \text{CH}_2\text{O}(g) + \text{H}_2(g) \rightarrow \text{CH}_4\text{O}(l) \quad \Delta H = -81.2 \text{ kJ} \][/tex]

To find the ΔH for the target reaction, we will need to manipulate these given reactions.

### Step-by-Step Solution:

1. Reverse the first reaction to align N[tex]\(_2\)[/tex]H[tex]\(_4\)[/tex] as a reactant:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
When a reaction is reversed, the sign of ΔH changes:
[tex]\[ \Delta H = -22.5 \text{ kJ} \][/tex]

2. Combine the reversed first reaction with the second reaction:
We can combine the reversed first reaction and the second reaction directly:

[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \quad \Delta H = -22.5 \text{ kJ} \][/tex]
[tex]\[ 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta H = 57.5 \text{ kJ} \][/tex]
Adding these two reactions:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) + 2\text{NH}_3(g) \rightarrow 2\text{NH}_3(g) + \text{N}_2(g) + 3\text{H}_2(g) \][/tex]
Cancel out the common terms ([tex]\(2\text{NH}_3(g)\)[/tex]):
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \][/tex]
The combined ΔH for these reactions is:
[tex]\[ \Delta H = -22.5 \text{ kJ} + 57.5 \text{ kJ} = 35.0 \text{ kJ} \][/tex]

3. Reverse the third reaction to align CH[tex]\(_2\)[/tex]O as a product and CH[tex]\(_4\)[/tex]O as a reactant:
[tex]\[ \text{CH}_4\text{O}(l) \rightarrow \text{CH}_2\text{O}(g) + \text{H}_2(g) \][/tex]
When a reaction is reversed, the sign of ΔH changes:
[tex]\[ \Delta H = 81.2 \text{ kJ} \][/tex]

4. Combine the result from step 2 with the reversed third reaction:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta H = 35.0 \text{ kJ} \][/tex]
[tex]\[ \text{CH}_4\text{O}(l) \rightarrow \text{CH}_2\text{O}(g) + \text{H}_2(g) \quad \Delta H = 81.2 \text{ kJ} \][/tex]
Adding these two reactions:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) + \text{CH}_4\text{O}(l) \rightarrow \text{N}_2(g) + 3\text{H}_2(g) + \text{CH}_2\text{O}(g) + \text{H}_2(g) \][/tex]
Cancel out the common terms ([tex]\(\text{H}_2(g)\)[/tex]):
[tex]\[ \text{N}_2\text{H}_4(l) + \text{CH}_4\text{O}(l) \rightarrow \text{CH}_2\text{O}(g) + \text{N}_2(g) + 3\text{H}_2(g) \][/tex]
The combined ΔH for these reactions is:
[tex]\[ \Delta H = 35.0 \text{ kJ} + 81.2 \text{ kJ} = -46.2 \text{ kJ} \][/tex]

Hence, the enthalpy change (ΔH) for the given reaction:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{CH}_4\text{O}(l) \rightarrow \text{CH}_2\text{O}(g) + \text{N}_2(g) + 3\text{H}_2(g) \][/tex]
is [tex]\(\Delta H = -46.2 \text{ kJ}\)[/tex].

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